$ E = \left[\begin{array}{rrr}-1 & 4 & 1 \\ 1 & -2 & 3\end{array}\right]$ $ A = \left[\begin{array}{rr}2 & 2 \\ -1 & 2 \\ 2 & -2\end{array}\right]$ What is $ E A$ ?
Solution: Because $ E$ has dimensions $(2\times3)$ and $ A$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ E A = \left[\begin{array}{rrr}{-1} & {4} & {1} \\ {1} & {-2} & {3}\end{array}\right] \left[\begin{array}{rr}{2} & \color{#DF0030}{2} \\ {-1} & \color{#DF0030}{2} \\ {2} & \color{#DF0030}{-2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{2}+{4}\cdot{-1}+{1}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{2}+{4}\cdot{-1}+{1}\cdot{2} & ? \\ {1}\cdot{2}+{-2}\cdot{-1}+{3}\cdot{2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{2}+{4}\cdot{-1}+{1}\cdot{2} & {-1}\cdot\color{#DF0030}{2}+{4}\cdot\color{#DF0030}{2}+{1}\cdot\color{#DF0030}{-2} \\ {1}\cdot{2}+{-2}\cdot{-1}+{3}\cdot{2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{2}+{4}\cdot{-1}+{1}\cdot{2} & {-1}\cdot\color{#DF0030}{2}+{4}\cdot\color{#DF0030}{2}+{1}\cdot\color{#DF0030}{-2} \\ {1}\cdot{2}+{-2}\cdot{-1}+{3}\cdot{2} & {1}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{2}+{3}\cdot\color{#DF0030}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-4 & 4 \\ 10 & -8\end{array}\right] $